Transportation Method of LP

 

Transportation and  Assignment Models 

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Introduction  

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In this chapter we will explore two special linear programming models 

The transportation model

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The assignment model

Because of their structure, they can be solved more efficiently than the simplex method These problems are members of a category of LP techniques called network  flow problems 

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10 – 2

 

Introduction  

Transportation model 

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The transportation problem deals with the distribution of goods from several points of ) to a number of points of supply (sources  demand (destinations  ) Usually we are given the capacity of goods at each source and the requirements at each destination Typically the objective is to minimize total transportation and production costs

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10 – 3

 

Introduction  

Example of a transportation problem in a network format Factories

Warehouses

100 Units

(Sources) Des Moines

(Destinations) Albuquerque

300 Units

300 Units

Evansville

Boston

200 Units

300 Units

Fort Lauderdale

Cleveland

200 Units

Capacities

Shipping Routes

Requirements 

Figure 10.1 Prentice-Hall, Inc.  © 2009 Prentice-Hall,

10 – 4

 

Introduction  

Assignment model 

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The assignment problem refers to the class of LP problems that involve determining the most efficient assignment of resources to tasks The objective is most often to minimize total costs or total time to perform the tasks at hand One important characteristic of assignment problems is one that machine only one or jobproject or worker can be assigned to

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10 – 5

 

Introduction  

Special-purpose algorithms 

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Although standard LP methods can be used to solve transportation and assignment problems, special-purpose algorithms have been developed that are more efficient They still involve finding and initial solution and developing improved solutions until an optimal solution is reached They are fairly simple in terms of computation

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10 – 6

 

Introduction  

Streamlined versions of the simplex method are important for two reasons 1. Their computation times are generally 100 times faster 2. They require less computer memory (and hence can permit larger problems to be solved)

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Two common techniques for developing initial solutions are the northwest corner method and Vogel’s approximation 

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The initial solution is evaluated using either the stepping-stone method or the modified distribution (MODI) method We also introduce a solution procedure called the Hungarian method , Flood’s technique, or the reduced matrix method 

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10 – 7

 

Setting Up a Transportation Problem  

The Executive Furniture Corporation manufactures office desks at three locations: Des Moines, Evansville, and Fort Lauderdale

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The firm distributes the desks through regional warehouses located in Boston, Albuquerque, and Cleveland Estimates of the monthly production capacity of each factory and the desks needed at each warehouse are shown in Figure 10.1

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10 – 8

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Setting Up a Transportation Problem  

Production costs are the same at the three factories so the only relevant costs are shipping from each source to each destination 

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Costs are constant no matter the quantity shipped The transportation problem can be described as how to select the shipping routes to be used and 

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the number of desks totransportation be shipped oncost  each route  so as to minimize total Restrictions regarding factory capacities and warehouse requirements must be observed

10  8

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Setting Up a Transportation Problem  



The first step is setting up the transportation table Its purpose is to summarize all the relevant data and keep track of algorithm computations Transportation costs per desk for Executive Furniture TO FROM DES MOINES

ALBUQUERQUE $5

BOSTON $4

CLEVELAND $3

EVANSVILLE

$8

$4

$3

FORT LAUDERDALE

$9

$7

$5

Table 10.1

10  9

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Setting Up a Transportation Problem  

Geographical locations of Executive Furniture’s factories and warehouses Boston

Cleveland

Factory

Des Moines Evanston

Warehouse

Albuquerque

Fort Lauderdale

Figure 10.2

10  10

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10 – 11

 

Setting Up a Transportation Problem  

Transportation table for Executive Furniture WAREHOUSE AT ALBUQUERQUE $5

WAREHOUSE AT BOSTON $4

WAREHOUSE AT CLEVELAND $3

EVANSVILLE FACTORY

$8

$4

$3

FORT LAUDERDALE FACTORY

$9

$7

$5

TO FROM DES MOINES FACTORY

WAREHOUSE REQUIREMENTS

300

200

200

Table 10.2 Cost of shipping 1 unit from Cleveland Fort Lauderdale factory to warehouse Boston warehouse demand

T otaldemand s supply upply and

Des Moines capacity constraint

FACTORY CAPACITY 100

300

300

700 Cell representing a source-to-destination (Evansville to Cleveland) shipping assignment that could be made

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Setting Up a Transportation Problem  

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In this table, total factory supply exactly equals total warehouse demand When equal demand and supply occur, a balanced problem is said to exist This is uncommon in the real world and we have techniques to deal with unbalanced problems

10 – 12

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Developing an Initial Solution:  Northwest Corner Rule  

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Once we have arranged the data in a table, we must establish an initial feasible solution One systematic approach is known as the northwest corner rule  Start in the upper left-hand cell and allocate units to shipping routes as follows 1. Exhaust the supply (factory capacity) of each row before moving down to the next row 2. Exhaust the demand (warehouse) requirements of each column before moving to the right to the next column 3. Check that all supply and demand requirements are met.

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In this problem it takes five steps to make the initial shipping assignments

10 – 13

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Developing an Initial Solution:  Northwest Corner Rule  1. Beginning in the upper left hand corner, we assign 100 units from Des Moines to Albuquerque. This exhaust the supply from Des Moines but leaves Albuquerque 200 desks short. We move to the second row in the same column. TO FROM

ALBUQUERQUE (A)

BOSTON (B )

CLEVELAND (C )

$5

$4

$3

EVANSVILLE (E )

$8

$4

$3

FORT LAUDERDALE (F )

$9

$7

$5

DES MOINES (D )

WAREHOUSE REQUIREMENTS

100

300

200

200

FACTORY CAPACITY 100

300

300

700

10 – 14

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Developing an Initial Solution:  Northwest Corner Rule  2. Assign 200 units from Evansville to Albuquerque. This meets Albuquerque’s demand. Evansville has 100 units remaining so we move to the right to the next column of the second row. TO FROM

ALBUQUERQUE (A)

DES MOINES (D )

100

EVANSVILLE (E )

200

FORT LAUDERDALE (F ) WAREHOUSE REQUIREMENTS

300

BOSTON (B )

CLEVELAND (C )

$5

$4

$3

$8

$4

$3

$9

$7

$5

200

200

FACTORY CAPACITY 100

300

300

700

10 – 15

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Developing an Initial Solution:  Northwest Corner Rule  3. Assign 100 units from Evansville to Boston. The Evansville supply has now been exhausted but Boston is still 100 units short. We move down dow n vertically to the next row in the Boston column. TO FROM

ALBUQUERQUE (A)

DES MOINES (D )

100

EVANSVILLE (E )

200

$5

$8

100

$9

FORT LAUDERDALE (F ) WAREHOUSE REQUIREMENTS

BOSTON (B )

300

200

CLEVELAND (C )

$4

$3

$4

$3

$7

$5

200

FACTORY CAPACITY 100

300

300

700

10 – 16

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Developing an Initial Solution:  Northwest Corner Rule  4. Assign 100 units from Fort Lauderdale to Boston. This fulfills Boston’s demand and Fort Lauderdale still has 200 units available.

TO FROM

ALBUQUERQUE (A)

DES MOINES (D )

100

EVANSVILLE (E )

200

WAREHOUSE REQUIREMENTS

$5

$8

$9

FORT LAUDERDALE (F )

300

BOSTON (B )

100

100

200

CLEVELAND (C )

$4

$3

$4

$3

$7

$5

200

FACTORY CAPACITY 100

300

300

700

10 – 17

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Developing an Initial Solution:  Northwest Corner Rule  5. Assign 200 units from Fort Lauderdale to Cleveland. This exhausts Fort Lauderdale’s supply and Cleveland’s demand. The initial shipment schedule is now complete. Table 10.3 TO FROM DES MOINES (D ) EVANSVILLE (E )

ALBUQUERQUE (A) 100

200

WAREHOUSE REQUIREMENTS

300

CLEVELAND (C )

$5

$4

$3

$8

$4

$3

$9

FORT LAUDERDALE (F )

BOSTON (B )

100

100

200

$7

200

200

$5

FACTORY CAPACITY 100

300

300

700

10 – 18

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Developing an Initial Solution:  Northwest Corner Rule  

We can easily compute the cost of this shipping assignment ROUTE

UNITS SHIPPED 100

PER UNIT x COST ($) 5

=

TOTAL COST ($) 500

FROM D

TO A

E

A

200

8

1,600

E

B 

100

4

400

F F

B  C 

100 200

7 5

700 1,000 4,200

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This solution is feasible but we need to check to see if it is optimal

10 – 19

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Stepping-Stone Method:  Stepping-Stone Finding a Least Cost Solution  

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The stepping-stone method is an iterative technique for moving from an initial feasible solution to an optimal feasible solution There are two distinct parts to the process 

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Testing the current solution to determine if improvement is possible Making changes to the current solution to obtain an improved solution

This process continues until the optimal solution is reached

10 – 20

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Stepping-Stone Method:  Stepping-Stone Finding a Least Cost Solution   

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There is one very important rule The number of occupied routes (or squares) must  always be equal to one less than the sum of the  number of rows plus the number ofthis columns  In the Executive Furniture problem means the initial solution must have 3 + 3 – 1 = 5 squares used Occupied shipping = Number of  – 1 routes (squares) of rows + Number columns

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When the number of occupied rows is less than this, the solution is called degenerate 

10 – 21

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Testing the Solution for  Possible Improvement  

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The stepping-stone method works by testing each unused square in the transportation table to see what would happen to total shipping costs if one unit of the product were tentatively shipped on an unused route There are five steps in the process

10 – 22

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  1. Select an unused square to evaluate 2. Beginning at this square, trace a closed path back to the original square via squares that are currently being used with only horizontal or vertical moves allowed 3. Beginning with a plus (+) sign at the unused square, place alternate minus ( –  –) signs and plus signs on each corner square of the closed path just traced

10 – 23

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  4. Calculate an improvement index by adding together the unit cost figures found in each square containing a plus sign and then subtracting unit sign costs in each square containing athe minus 5. Repeat steps 1 to 4 until an improvement index has been calculated for all unused squares. If all indices computed are greater than or equal to zero, an optimal solution has been reached. If not, it is possible to improve the current solution and decrease total shipping costs.

10 – 24

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  

For the Executive Furniture Corporation data

Steps 1 and 2 . Beginning with Des Moines –Boston route we trace a closed path using only currently occupied squares, alternately placing plus and minus signs in the corners of the path 

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In a closed path , only squares currently used for shipping be used corners Only one can closed routein is turning possible for each square we wish to test

10 – 25

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  Step 3 . We want to test the cost-effectiveness of the Des Moines –Boston shipping route so we pretend we are shipping one desk from Des Moines to Boston and put a plus in that box  But if we ship one more unit out of Des Moines we will be sending out 101 units  Since the Des Moines factory capacity is only 100, we must ship fewer desks from Des Moines



to Albuquerque Albuquerque so we place a minus sign in that box But that leaves Albuquerque one unit short so we must increase the shipment from Evansville to Albuquerque by oneclosed unit and so on until we we complete the entire path

10 – 26

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  

Evaluating the unused Des Moines –Boston shipping route

Warehouse A Factory D 

Factory E 

TO FROM

ALBUQUERQUE

BOSTON

CLEVELAND

$5

$4

$3

$4

$3

DES MOINES

100

EVANSVILLE

200

$9

FORT LAUDERDALE WAREHOUSE REQUIREMENTS

$8

300

100

100

200

$7

200

200

$5

Warehouse B 

$5 100

 – 

+

$4

+ $8

200

 –  $4 100

FACTORY CAPACITY 100

300

300

700

Table 10.4

10 – 27

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  

Evaluating the unused Des Moines –Boston shipping route

Warehouse A Factory D 

Factory E 

TO FROM

ALBUQUERQUE

BOSTON

CLEVELAND

$5

$4

$3

$4

$3

DES MOINES

100

EVANSVILLE

200

$9

FORT LAUDERDALE WAREHOUSE REQUIREMENTS

$8

300

100

100

200

$7

200

200

$5

Warehouse B 

$5

99

$4 1

100

 – 

201

+

+ $8

200

99

 –  $4 100

FACTORY CAPACITY 100

300

300

700

Table 10.4

10 – 28

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  

Evaluating the unused Des Moines –Boston shipping route

Warehouse A Factory D 

Factory E 

TO FROM

ALBUQUERQUE

BOSTON

CLEVELAND

$5

$4

$3

$4

$3

DES MOINES

100

EVANSVILLE

200

$9

FORT LAUDERDALE WAREHOUSE REQUIREMENTS

$8

300

100

100

200

$7

200

200

$5

Warehouse B 

$5

99

$4 1

100

 – 

201

+

+ $8

200

FACTORY Result CAPACITY

99

 –  $4 100

of Proposed

Shift in Allocation 100 = 1 x $4  – 1 x $5 300 + 1 x $8  – 1 x $4 = +$3 300

700

Table 10.4

10 – 29

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  Step 4 . We can now compute an improvement index   ( I ij) for the Des Moines –Boston route 

We add the costs in the squares with w ith plus signs and subtract the costs in the squares with minus signs Des Moines – Boston index =  I  DB = +$4 – $5 + $5 – $4 = + $3

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This means for every desk shipped via the Des Moines –Boston route, total transportation cost will increase by $3 over their current level

10 – 30

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  Step 5 . We can now examine the Des Moines – Cleveland unused route which is slightly more difficult to draw 

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Again we can only turn corners at squares that represent existing routes We must pass through the Evansville –Cleveland square but we can not turn there or put a + or –  sign The closed path we will use is + DC   – – DA + EA –  – EB + FB  –  – FC 

10 – 31

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  

Evaluating the Des Moines –Cleveland shipping route TO

FROM DES MOINES

EVANSVILLE

ALBUQUERQUE

BOSTON

$5

$4

 – 

+

100

200

WAREHOUSE REQUIREMENTS

300

Start

$3

+ $8

100

$4

$3

 – 

$9

FORT LAUDERDALE

CLEVELAND

100

+

200

$7

200

 – 

200

$5

FACTORY CAPACITY 100

300

300 700

Table 10.5

Des Moines –Cleveland improvement index =  I  DC  = + $3 – $5 + $8 – $4 + $7 – $5 = + $4

10 – 32

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Five Steps to Test Unused Squares  with the Stepping-Stone Method  

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Opening the Des Moines –Cleveland route will not lower our total shipping costs Evaluating the other two routes we find EvansvilleCleveland index =  I  EC  = + $3 – $4 + $7 – $5 = + $1 

The closed path is  – EB + FB  – + EC  –  – FC 

 – Fort Lauderdale =  I  F  FA A = + $9 – $7 + $4 – $8 = – $2 Albuquerque index 

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The closed path is  – FB + EB  –  – EA + FA –

So opening the Fort Lauderdale-Albuquerque route will lower our total transportation costs

10 – 33

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Obtaining an Improved Solution  

In the Executive Furniture problem there is only one unused route with a negative index (Fort Lauderdale-Albuquerque)

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If there was more than one route with a negative index, we would choose the one with the largest improvement We now want to ship the maximum allowable

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

number of units on the new route The quantity to ship is found by referring to the closed path of plus and minus signs for the new route and selecting the smallest number found in those squares containing minus signs

10 – 34

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Obtaining an Improved Solution  

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To obtain a new solution, that number is added to to all squares on the closed path with w ith plus signs and subtracted from all squares the closed path with minus signs All other squares are unchanged In this case, the maximum number that can be shipped is 100 desks as this is the smallest value in a box with a negative sign (FB route) We add 100 units to the FA and EB routes and subtract 100 from FB and EA routes This leaves balanced rows and columns and an improved solution 

10 – 35

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Obtaining an Improved Solution  

Stepping-stone path used to evaluate route FA TO

A

FROM D 

100

E 

200

Table 10 10.6 .6

$5

$8

+

 – 

$9

F 

WAREHOUSE REQUIREMENTS

B

+

300

100

100

 –  200

FACTORY CAPACITY

C 

$4

$3

$4

$3

$7

200 200

$5

100

300

300 700

10 – 36

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Obtaining an Improved Solution  

Second solution to the Executive Furniture problem  TO

FROM

A

D 

100

E 

100

F 

100

WAREHOUSE REQUIREMENTS

300

B

$5

$8

200

$9

$4

$3

$4

$3

$7

200

FACTORY CAPACITY

C 

200 200

$5

100

300

300 700

Table 10 10.7 .7 

Total shipping costs have been reduced by (100 units) x ($2 saved per unit) and now equals $4,000

10 – 37

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Obtaining an Improved Solution   

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This second solution may or may not be optimal To determine whether further improvement is possible, we return to the first five steps to test each square is now unused The four newthat improvement indices are  D to B =  I  DB = + $4 – $5 + $8 – $4 = + $3 (closed path: + DB  –  – DA + EA –  – EB )  D to C =  I  DC  = + $3 – $5 + $9 – $5 = + $2  – DA + FA –  – FC )  (closed path: + DC  – E to C =  I  EC  = + $3 – $8 + $9 – $5 = – $1  – EA + FA –  – FC )  (closed path: + EC  – F to B =  I  FB = + $7 – $4 + $8 – $9 = + $2  – EB + EA – (closed path: + FB  –  – FA) 

10 – 38

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Obtaining an Improved Solution  

Path to evaluate for the EC route TO

A

FROM D 

100

E 

100

F 

WAREHOUSE REQUIREMENTS

B

$5 $8

 – 

100

+

300

$4

200

$9

$4

$3

Start

$3

+ $7

200

FACTORY CAPACITY

C 

200

 – 

200

$5

100

300

300 700

Table 10 10.8 .8 

An improvement can be made by shipping the maximum allowable number of units from E to C 

10 – 39

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Obtaining an Improved Solution  

Total cost of third solution ROUTE

DESKS SHIPPED

PER UNIT x COST ($)

=

TOTAL COST ($)

FROM

TO

D

A

100

5

500

E

B 

200

4

800

E

C 

100

3

300

F

A

200

9

1,800

F

C 

100

5

500 3,900

10 – 40

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Obtaining an Improved Solution  

Third and optimal solution TO

FROM D 

A

100

WAREHOUSE REQUIREMENTS

Table 10 10.9 .9

$5 $8

E 

F 

B

200 300

$4

200

$9

$4

$7

200

FACTORY CAPACITY

C 

$3

100

100 200

$3

$5

100

300

300 700

10 – 41

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Obtaining an Improved Solution  

This solution is optimal as the improvement indices that can be computed are all greater than t han or equal to zero D to B =  I  DB = + $4 – $5 + $9 – $5 + $3 – $4 = + $2 (closed path: + DB  –  – DA + FA –  – FC + EC  –  – EB )  D to C =  I  DC  = + $3 – $5 + $9 – $5 = + $2 (closed path: + DC  –  – DA + FA –  – FC )  E to A =  I  EA = + $8 – $9 + $5 – $3 = + $1  – FA + FC  –  – EC )  (closed path: + EA – F to B =  I  FB = + $7 – $5 + $3 – $4 = + $1  – EB )  (closed path: + FB  –  – FC + EC  –

10 – 42

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Summary of Steps in Transportation  Algorithm (Minimization)  1. Set up a balanced transportation table 2. Develop initial solution using either the northwest corner method or Vogel’s approximation method  3. Calculate an improvement index for each empty cell using either the stepping-stone method or the MODI method. If improvement indices are all nonnegative, stop as the optimal solution has been4. found. If any index is negative, continue to step 4. Select the cell with the improvement index indicating the greatest decrease in cost. Fill this cell using the stepping-stone path and go to step 3.

10 – 43

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Using Excel QM to Solve  Transportation Problems  

Excel QM input screen and formulas

Program 10.1A

10 – 44

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Using Excel QM to Solve  Transportation Problems  

Output from Excel QM with w ith optimal solution

Program 10.1B

10 – 45

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Vogel’s Approximation Method: Another Way To Find An Initial Solution  

Vogel’s Approximation Method (VAM ) is not as

simple as the northwest corner method, but it provides a very good initial solution, often one 

that is the optimal solution VAM tackles the problem of finding a good initial solution by taking into account the costs associated with each route alternative





This something that the northwest corner rule doesis not do To apply VAM, we first compute for f or each row and column the penalty faced if we should ship over the second-best route instead of the least-cost   route

10 – 46

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Vogel’s Approximation Method   

The six steps involved in determining an initial VAM solution are illustrated below beginning with the same layout originally shown in Table 10.2 VAM Step 1. For each row and column of the transportation table, find the difference between the distribution cost on the best route in the row or column and the second best route in the row or

column  This is the opportunity cost of not using the best route  Step 1 has been done in Table 10.11

10 – 47

 © 2009 Prentice-Hall, Prentice-Hall, Inc.  

Vogel’s Approximation Method   

Transportation table with VAM row and column differences shown

TO FROM

D  E 

3 

0 

0 

A

B

C 

100

TOTAL REQUIRED

Table 10.11

$4

$3

$8

$4

$3

100 $9

300

TOTAL AVAILABLE

$5

200

F 

OPPORTUNITY COSTS

100 200

$7

200 200

$5

100

1

300

1

300

2

700

10 – 48

 © 2009 Prentice-Hall, Prentice-Hall, Inc.  

Vogel’s Approximation Method   VAM Step 2 . identify the row or column with the greatest opportunity cost, or difference (column A in this example) VAM Step 3 .Assign .Assign as many units as possible to the lowest-cost square in the row or column selected VAM Step 4 . Eliminate any row or column that has been completely satisfied by the assignment just

made by placing X s in each appropriate square VAM Step 5 . Recompute the cost differences for the transportation table, omitting rows or columns eliminated in the previous step

10 – 49

 © 2009 Prentice-Hall, Prentice-Hall, Inc.  

Vogel’s Approximation Method   

VAM assignment with D ’s ’s requirements satisfied 

TO FROM

D 

3 1 

0 3 

0 2 

A

B

C 

100

$5

$4

X

$8

OPPORTUNITY COSTS TOTAL AVAILABLE $3

X

$4

$9

TOTAL REQUIRED

Table 10.12

300

$7

200

1

300

1

300

2

$3

E  F 

100

$5

200

700

10 – 50

 © 2009 Prentice-Hall, Prentice-Hall, Inc.  

Vogel’s Approximation Method   VAM Step 6 . Return to step 2 for the rows and columns remaining and repeat the steps until an initial feasible solution has been obtained 







In this case column B now has the greatest difference, 3 We assign 200 units to the lowest-cost square in the column, EB the differences and find the We recompute greatest difference is now in row E  We assign 100 units to the lowest-cost square in the column, EC 

10 – 51

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Vogel’s Approximation Method   

Second VAM assignment with B ’s ’s requirements satisfied OPPORTUNITY TO FROM

D 

3 1  A 100

0 3  B $5

$8

$4

X

200

COSTS

0 2  C 

TOTAL AVAILABLE $3

X

$4

$3

100

300

1

E  $9

F  TOTAL REQUIRED

Table 1 10.13 0.13

1

300

$7

X 200

$5

200

300 700

2

10 – 52

 © 2009 Prentice-Hall, Prentice-Hall, Inc.  

Vogel’s Approximation Method   

Third VAM assignment with E ’s ’s requirements satisfied TO

A

FROM

D 

100

E 

X

Table 10.14

$5

$8

$9

F  TOTAL REQUIRED

B

300

$4

X

200

$4

$3

X

100

$7

X 200

TOTAL AVAILABLE

C 

$3

$5

200

100

300

300 700

10 – 53

 © 2009 Prentice-Hall, Prentice-Hall, Inc.  

Vogel’s Approximation Method   

Final assignments to balance column and row requirements TO

A

FROM

D 

100

E 

X

F 

200

TOTAL REQUIRED

Table 10.15

B $5

$8

300

$9

$4

X

200

$4

$7

X 200

TOTAL AVAILABLE

C  $3

X

100

100 200

$3

$5

100

300

300 700

10 – 54